Is there a rule that states that I should not divide in this scenario?

[Callmelucky]

Homework Statement

Write the coordinates of points where $f(x)=\log _{2} x^2 -1$ intersects x and y axis

Relevant Equations

$\log _{a} x^n = n\times \log _{a} x$

So basically this is how I solved this problem:
1. $f(x)=\log _{2} x^2 - 1$
2. $0=\log _{2} x^2 -1$
3. $1= 2\times \log _{2} x$
4. $\frac{1}{2}= \log _{2} x$
5. $2^{\frac{1}{2}}=x=\sqrt{2}$

So I wrote coordinates to be ($\sqrt{2}$, 0)

But apparently, that is not the only solution. There should be another answer with a negative sign so ($\pm\sqrt{2}$, 0) would be a complete solution for points at which graphs cross the x line. There are no points where the graph crosses the y-line.
This is how it's solved in the textbook(pic in attachments). And I understand that it's correct because the graph really does cross the x-line in those points.
So, my question is, is there a rule I am not aware of that states that I can't divide an equation with n(exponent of an argument moved down to the front)?

Thank you.

 

[weirdoguy]

What is the domain of your function? The problem is basically in the 3rd step.

 

[Callmelucky]

What is the domain of your function? The problem is basically in the 3rd step.

I didn't learn about domain, in Croatia we learn that in 4th grade of high school, I am in 2nd right now. Are you saying that I can't move 2 down if it's original place is up?

 

[weirdoguy]

You can, but it will be $2\log_2|x|$ instead of what you wrote. That's because your $x$'s can be negative, eg: for $x=-2$ we have $\log_2(-2)^2=\log_24=2$

 

[Callmelucky]

You can, but it will be $2\log_2|x|$, because your $x$'s can be negative.

Ohh, that make sense, should I always do that when moving 2 down? Because I haven't done that once so far

 

[weirdoguy]

In the cases of functions like this one in your example yes. The rule: $\log a^n=n\log a$ works only for positive $a$, and in the function you gave your "$a$" (which is $x$) could be negative.

 

[Callmelucky]

In the cases of functions like this one in your example yes. The rule: $\log a^n=n\log a$ works only for positive $a$, and in the function you gave your "$a$" (which is $x$) could be negative.

Okay, thank you very much. :D

 

[Mark44]

Homework Statement: Write the coordinates of points where $f(x)=\log _{2} x^2 -1$ intersects x and y axis
Relevant Equations: $\log _{a} x^n = n\times \log _{a} x$

So basically this is how I solved this problem:
1. $f(x)=\log _{2} x^2 - 1$
2. $0=\log _{2} x^2 -1$
3. $1= 2\times \log _{2} x$
4. $\frac{1}{2}= \log _{2} x$
5. $2^{\frac{1}{2}}=x=\sqrt{2}$

So I wrote coordinates to be ($\sqrt{2}$, 0)

From line 2, you have $\log_2(x^2) = 1 \Rightarrow x^2 = 2 \Rightarrow x= \pm \sqrt 2$

But apparently, that is not the only solution. There should be another answer with a negative sign so ($\pm\sqrt{2}$, 0) would be a complete solution for points at which graphs cross the x line. There are no points where the graph crosses the y-line.

 

[Callmelucky]

From line 2, you have $\log_2(x^2) = 1 \Rightarrow x^2 = 2 \Rightarrow x= \pm \sqrt 2$

yeah, but I was confused why the method I was using so far isn't working anymore. Thank you.

 

[Mark44]

yeah, but I was confused why the method I was using so far isn't working anymore. Thank you.

It's because your relevant equation -- $\log_a (x^n) = n\log_a(x)$ -- is valid only for x > 0. Although $x = \sqrt 2$ is a solution of equations 2 and 4, $x = -\sqrt 2$ is also a solution of the original equation.

 

[fresh_42]

yeah, but I was confused why the method I was using so far isn't working anymore. Thank you.

You have to use $\log_2 x^2 =2\cdot \log |x|$ since the logarithm doesn't allow negative arguments. Therefore you end up with $2^{1/2}=\sqrt{2}=|x|$ which includes both signs!